Problem: Perform the row operation, $-4R_2\rightarrow R_2$, on the following matrix. $\left[\begin{array} {ccc} 5 & 7 & 8 & 5 \\ 4 & 1 & 4 & 7 \\ 2 & 0 & 2 & 1 \end{array} \right] $
Answer: Background There are three basic row operations that can be performed on matrices. $R_i \leftrightarrow R_j$. This symbol tells us to interchange rows $i$ and $j$. $cR_i \rightarrow R_i$. This symbol tells us to multiply a row $i$ by a constant $c$. $R_i + cR_j \rightarrow R_i$. This symbol tells us to add $c$ times row $j$ to row $i$. Finding the new row to be used For the given matrix, $R_2$ is given below. $R_2=\left[\begin{array} {ccc} 4 & 1 & 4 & 7 \end{array} \right]$ We are asked to perform the row operation, $-4R_2\rightarrow R_2$. Therefore, we must multiply $R_2$ by $-4$. $\begin{aligned}-4R_2 &= -4\left[\begin{array} {ccc} 4 & 1 & 4 & 7 \end{array} \right] \\\\&=\left[\begin{array} {ccc} -16 & -4 & -16 & -28 \end{array} \right]\end{aligned}$ Substituting the row Now, we must substitute row $R_2$ with $-4R_2$. $\left[\begin{array} {ccc} 5 & 7 & 8 & 5 \\ {4} & {1} & {4} & {7} \\ 2 & 0 & 2 & 1 \end{array} \right] \xrightarrow{-4R_2\rightarrow R_2} \left[\begin{array} {ccc} 5 & 7 & 8 & 5 \\ {-16} & {-4} & {-16} & {-28} \\ 2 & 0 & 2 & 1 \end{array} \right] $ Summary Our resultant matrix is the following. $\left[\begin{array} {ccc} 5 & 7 & 8 & 5 \\ -16 & -4 & -16 & -28 \\ 2 & 0 & 2 & 1 \end{array} \right]$